2009 Zhautykov International Olympiad Problems/Problem 3
First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.
Let the incircle of touch , , at , , respectively. By Ceva's theorem, it is easy to see that , , are concurrent at a point , which we call the Gergonne point of .
Let , , . As shown in the figure, by Stewart's theorem, we have In , by Menelaus' theorem, we have Solving this, we get
Noting that by AM-GM inequality, we have
Lemma: Let be the Gergonne point of , then
Returning to the original problem, let be the Gergonne point of . As shown in the figure, by the lemma, we have Similarly, we have , , adding these together, we get the desired result.